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2 years ago

Please help me! I am very confused

Mathematics

1 answer:

Sphinxa [80]2 years ago

4 0

Answer:

1) 22.5 A

2) 112 B

3) 430 L

4) 576 P

5) 486 L

6) 624 S

Code is A B L P L S

Step-by-step explanation:

1) Area of a triangle = 1/2 * base * height = 1/2 *5*9 = 22.5

2) Area of a parallelogram = base x height = 14 x 8 = 112

3) Area of a rectangular prism = 2(length * width) + 2*(length + height) + 2 *(length *height)

= 2(15x7 + 15*5 + 7*5)

= 2(105 + 75 + 35)

= 2 * 215

= 430

4) Volume of a triangular prism = 1/2 * base * length * height

= 1/2 * 8 * 16 * 9

= 576

5) A cube has six surfaces. Each surface has an area of s x s where s is the length of each side. In this case, each side has area of 9x9 = 81. Total surface area = 6 x 81 = 486 and that is the paper required

6) The trailer is a rectangular prism so its volume = length x width x height = 13 6 x 8 = 624

Now you have to look at each value and see which letter it corresponds to. For example answer 1) is 22.5 which lies between 0-100 so it gets letter A, answer (2) is 112 which lies in the range 101-200 so it gets the letter B and so on

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Two trains start from Fort Worth traveling at the same speed. The trip for Train W takes 20 hours, while the trip for Train Z ta

MrMuchimi

Answer:

The equation that best models this situation is :

b. $\frac{x}{20}=\frac{x+400}{25}$

Distance traveled by Train W = 1600 miles

Distance traveled by Train Z = 2,000 miles

Step-by-step explanation:

Given:

Time taken for Train W to complete a trip = 20 hours

Time taken for Train Z to complete a trip = 25 hours

The city Train Z is traveling to is 400 miles farther away than the city Train W is traveling to.

both trains have same speeds and start from same location.

To find the distance in total each train travels.

Solution:

Let length of the trip of Train W be = $x$ miles

Speed of Train W can be given as :

⇒ $\frac{Distance}{Time}$

⇒ $\frac{x}{20}\ miles/h$

So, the length of the trip of Train Z will be = $(x+400)$ miles

Speed of Train Z can be given as :

⇒ $\frac{Distance}{Time}$

⇒ $\frac{x+400}{25}\ miles/h$

Since the speeds are same, so the equation to find $x$ can be given as:

⇒ $\frac{x}{20}=\frac{x+400}{25}$

Solving for $x$

Multiplying both sides by 100 to remove fractions.

⇒ $100\times \frac{x}{20}=100\times \frac{x+400}{25}$

⇒ $5x=4(x+400)$

Using distribution.

⇒ $5x=4x+1600$

Subtracting both sides by $4x$

⇒ $5x-4x=4x-4x+1600$

⇒ $x=1600$

Thus, Distance traveled by Train W = 1600 miles

Distance traveled by Train Z = $1600+400$ = 2,000 miles

3 0

3 years ago

I need some help with a math problem

alina1380 [7]

Answer:

3m-25>68

Step-by-step explanation:

Your Welcome

Btw- I didn't write a step by step explanation cause it doesn't need one.

4 0

2 years ago

Kelli and her family went to the beach for a vacation. They drove 293 miles in 7 hours to get there. ABOUT how many miles did th

12345 [234]

Answer:

41.86 miles each hour

Step-by-step explanation:

Kelli and her family went to the beach for a vacation. They drove 293 miles in 7 hours to get there. ABOUT how many miles did they drive each hour?

This Question is calculated as:

7 hours = 293 miles

1 hour = x miles

Cross Multiply

7 hours × x = 1 × 293 miles

x = 293 miles/7 hours

x = 41.857142857 miles per hour

x ≈ 41.86 miles per hour

Hence, they drove 41.86 miles each hour

4 0

2 years ago

The ratio of the number of dogs the number of cats is 4:5. There are 270 animals on the rescue Farm. The number of dogs is 4/5.

mafiozo [28]

Answer:

Here is the complete question:

The ratio of the number of dogs to the number of cats is 4:5. There are 270 animals on the rescue farm. for each of the following statements, explain whether statements is true or false and why:

a. The number of dogs is 4/5 the number of cats.

b. 4/5 of the pets at the farm are dogs.

c. There are exactly 30 more cats than dogs.

d. There are exactly 30 dogs at the farm.

e. 5/9 of the pets at the farm are cats.

a,c and e are true statements.

Step-by-step explanation:

Let the number of dogs be "d' and number of cats be "c".

According to the question:

⇒ $c+d=270$ ...equation (i)

Arranging the ratio:

⇒ $\frac{4}{5} = \frac{d}{c}$

⇒ $4c=5d$

⇒ $c=(\frac{5}{4})d$ ...equation (ii)

Plugging (ii) in equation (i).

⇒ $c+d=270$

⇒ $\frac{5}{4} d+d=270$

⇒ $\frac{5d+4d}{4} =270$

⇒ $9d=270\times 4$

⇒ $d=\frac{270\times 4}{9}$

⇒ $d=120$

Number of dogs "d" = 120

Number of cats "c" = (270-120) = 150

Lets check each statement:

The number of dogs is 4/5 the number of cats.

⇒ $\frac{4}{5} \times 150$

⇒ $\frac{4\times 150}{5}$

⇒ $120$

Statement is true.

4/5 of the pets at the farm are dogs.

⇒ $\frac{4}{5} \times 270$

⇒ $216$

Number of dogs = 120

So the above statement is false.

There are exactly 30 more cats than dogs.

⇒ $c-d =30$

⇒ $150-120=30$

Statement is true.

There are exactly 30 dogs at the farm.

False, as there are 120 dogs

5/9 of the pets at the farm are cats.

⇒ $\frac{5}{9} \times 270$

⇒ $150$

Statement is true.

So,

Statement a,c and e are true as we could observe in the explanation.

8 0

2 years ago

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp

olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = $\overline{x}$ = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data speeds i.e. x - $\overline{x}$

x - $\overline{x}$ = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between carrier's highest data speed and the mean of all 50 data speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

$\frac{59.98}{20.03}=2.99$

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

$z=\frac{x-\overline{x}}{s}$

Using the given values, we get:

$z=\frac{75.6-15.62}{20.03}=2.99$

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the carrier's highest data speed is significantly higher.

3 0

3 years ago